Netforce of spring, hookes law. help

[jim gerth]

Homework Statement

An "extreme" pogo stick utilizies a spring whose uncompressed length is 46 cm and whose force constant is 1.4 x 10^4 N/m (14000). A 60 kg enthusiast is jumping on the pogo stick, compressing the spring to a length of only 5.0 cm at the bottom of her jump, Calculate; a) The net upward force on her at the moment the spring reaches its greatest compression and b) her upward acceleration in m/s^2 and gs at that moment.

Any thoughts?
Thanks.

Homework Equations

F=-kx F=ma w=mg

The Attempt at a Solution

Ok so i started with multiplying the force constant and difference of distance . 14000 x (0.46 - 0.05) = 5740 then i subtracted the weight which is (60)(9.8)= 588 which gave me an answer of 5152. The answer in the book says 5100. I know with having 2 sig figs you could say it was 5100 but i am not sure if i went about this right.. and also don't know where to go with b especially if i did not get a correct.??

thanks everyone.

 

[Shinaolord]

Having 2 significant figures with your final result does NOT give 5100. It gives 5200.

$\vec{F}_{net} = \vec{F}_{spring} + \vec{F}_{person}$
Both of which you stated the formulas for.
If you round both of your results with respect to the amount of significant figures before subtracting, you get 5100 ( after significant figures are applied to this as well, beforehand it is 5110)
That's the only thing "wrong" I see.

Your analysis was correct, however. What are your thoughts on the second portion (the acceleration)?

 

[jim gerth]

Thanks for the reply and clarification, Yeah sorry , i meant to say 5200.. and i have no idea how to approach the acceleration , we have yet to learn that. I've been checking everywhere on how to approach it but i don't even know where to start lol -____- answers in the back is 85 m/s^2 and 8.7 gs. i wish i could say my answers were even close .

 

[jim gerth]

lol i thought it was more than just that but nvm i got it . Thanks again

 

[HalfLostAndSearching]

Hello,

First of all, great job on starting with the Hooke's Law equation and using the correct units for the force constant and distance.

For part a), you did the calculation correctly, but there is a slight error in your units. The force constant is in N/m, so the unit for the distance should be in meters as well. This will give you a net upward force of 5100 N, which is the same as the book's answer.

For part b), you can use the equation F=ma to find the acceleration, since you already have the net force. Just plug in the values for the force (5100 N) and the mass (60 kg) and solve for acceleration. Then, convert the answer to m/s^2 and gs using the appropriate conversion factors.

I hope this helps! Let me know if you have any other questions.